Question

Consider the following unbalanced equation:

$\mathit{C}{\mathit{a}}_{\mathbf{3}}{\mathbf{\left(}\mathbf{P}\mathbf{O}\mathbf{4}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\to }\mathit{C}\mathit{a}\mathit{S}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{3}}\mathit{P}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$

What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0kg of calcium phosphate with 1.0 kg of concentrated sulfuric acid (98% H₂SO₄ by mass)?

Short answer

The masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0kg of calcium phosphate with 1.0 kg of concentrated sulfuric acid is$1.3\times {10}^{3}g,6.3\times {10}^{2}g$

Step-by-step solution

Definition

The molar mass of a materialis defined as the mass of the substancecontained in one moleof the substance, and it isequal to the total massof the component atoms. It may be stated mathematically as follows:
Molar mass =$\frac{Weighting}{numberofmoles}$

Balancing the unbalanced

The unbalanced aka given equation is

$C{a}_3{\left(\mathrm{PO}4\right)}_2\left(s\right)+H_{2}S{O}_4\left(aq\right)\to CaS{O}_4\left(s\right)+H_{3}P{O}_4\left(aq\right)$

And the balanced equation,

$C{a}_3{\left(\mathrm{PO}4\right)}_2\left(s\right)+3H_{2}S{O}_4\left(aq\right)\to 3CaS{O}_4\left(s\right)+2H_{3}P{O}_4\left(aq\right)$

To figure out how many products can be made from a given amount of reactants, start by identifying the limiting reagent, which restricts the number of products that can be made.

Initially, we have 1.0 kilogram calcium phosphate and 1.0 kg concentrated sulfuric acid with a sulfuric acid content of 98 percent by mass.

As a result, the sulfuric acid's starting mass is

$$\begin{gathered} =\frac{98kg{H}_{2}S{O}_4}{100kgconc{H}_{2}S{O}_4}\times 1.0kgconc{H}_{2}S{O}_4 \\ =0.98kg{H}_{2}S{O}_4 \end{gathered}$$

Number of moles of calcium phosphate

Using their molar masses, convert the quantities (in mass) of calcium phosphate and sulfuric acid to moles.

$\begin{array}{rcl}1kgC{a}_3{\left(P{O}_4\right)}_2& =& \frac{1molCa}{}\end{array}$₃PO₄₂310.18×10^-3kgCa₃PO₄₂=3.2molCa3(PO4)2 0.98kgH2SO4=1molH2SO498.09×10-3kgH2SO4=10molH2So43.2molCa3(PO4)2=3molH₂SO₄1molCa₃(PO₄)₂=9.6molH2SO4

One mole of calcium phosphate interacts with three moles of sulfuric acid in a balanced chemical reaction. The amount of sulfuric acid required to thoroughly react with 3.2 moles of calcium phosphate.

Thus we have 10 mol H₂SO₄, so calcium phosphate is the limiting reagent

Masses of calcium phosphate and phosphoric acid produced

One mole of calcium phosphate interacts with three moles of sulfuric acid to generate three moles of calcium sulphate and two moles of phosphoric acid in a balanced chemical reaction.

Using the stoichiometric relationship, determine the amount of calcium sulphate and phosphoric acid generated from 3.2 mol Ca₃(PO₄)₂

$$\begin{gathered} 3.2molC{a}_3{\left(P{O}_4\right)}_2\times \frac{3molCaS{O}_4}{1molC{a}_3{\left(P{O}_4\right)}_2}\times \frac{136.15gCaS{O}_4}{1molCaS{O}_4}=1.3\times {10}^{3}gCaS{O}_4 \\ 3.2molC{a}_3{\left(P{O}_4\right)}_2\times \frac{2mol{H}_{3}P{O}_4}{1molC{a}_3{\left(P{O}_4\right)}_2}\times \frac{97.99g{H}_{3}P{O}_4}{1mol{H}_{3}P{O}_4}=6.3\times {10}^{2}g{H}_{3}P{O}_4 \end{gathered}$$

Therefore the masses of calcium phosphate and phosphoric acid produced is

$1.3\times {10}^{3}g,6.3\times {10}^{2}g$